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Time dilation and relative reference frames

  1. Nov 21, 2015 #1
    I understand that if someone is traveling away from earth at a very high speed, time will slow down for the traveler relative to the people on earth. However, why is it not the other way around? If there is no universal reference frame, could this situation not also be thought of as the traveler remaining stationary and the earth traveling away from the traveler at very high speed? In this case, time should slow down for people on earth relative to the traveler.
  2. jcsd
  3. Nov 21, 2015 #2


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    Yes, the situation is completely symmetric. Each says the other one's time is slow. You may find section 3.10 of my book Relativity for Poets helpful: http://lightandmatter.com/poets/
  4. Nov 21, 2015 #3
    That's correct, time is always measured to be slower on the "moving" clock compared to the "stationary" clock. "Moving" and "stationary" are in quotes because the choice of which is which is arbitrary. Your scenario is symmetrical and time dilation is mutual.

    Don't try to stretch this to the twins "paradox" scenario, that one is not symmetrical overall.
  5. Nov 21, 2015 #4
    So the conclusion of the thought experiment is that upon returning to earth, the traveler has aged only slightly, while his friends and family are much older. Why is it not the other way around - the traveler is older and the people on earth have barely aged?
  6. Nov 21, 2015 #5
    Re-read your original post and the replies you got. There is no returning to Earth anywhere in there, you described a symmetric scenario and your conclusion for it was correct.

    I already advised not stretching that conclusion to the twins "paradox" but you went ahead and did just that. It's not that simple there, that scenario is not symmetric. Search and read the countless threads on the twins "paradox" for more details.
    Sure, but you'd have to arrange for Earth to do the turning around while the "traveler" remains inertial. Who is moving or at rest is just a matter of choice of reference, but who does the turning around is not.
  7. Nov 21, 2015 #6


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    The short answer is that your wrist watch is measuring something rather like distance travelled through space-time, rather as the odometer in your car measures distance travelled through space. The traveller took a shorter route through space-time than the stay-at-home, so his wrist watch hasn't ticked as many times as the stay-at-home.

    Another way to look at it is that, in relativity, "now" is something that depends on your state of motion (among other things). You can't see what's happening "now" anywhere except exactly where you are because of the finite speed of light. In fact, things happening "now" but not where you are cannot affect you for a little while because they can't get to you faster than light, and this allows you some flexibility in the definition of "now". It turns out to be convenient to define "now" differently before and after any acceleration. One or other twin must accelerate if they are to meet up - otherwise they just float away from each other. When one accelerates, he changes his definition of "now", and it turns out that the difference between "now" just before he turns round and "now" just after he turns round accounts for the discrepancy.

    Clear as mud? An example may help. I stay at home for two years (by my watch); you go out-and-back at such a speed that the time dilation factor is 2. I say your watch is ticking slowly, and reads six months when you turn round and one year when you get back. But for you the situation is a bit more complex. You say my clock is ticking slowly, so that only three months have elapsed on my clock just before you turn round. Just after you turn round, you change your definition of "now" and say that twenty-one months have elapsed on my clock. Again, three months elapse on my clock "during" your journey home and you are unsurprised to see that my clock says two years have elapsed when you get home.

    That's only book-keeping, however. It's all hidden from you by the finite speed of light and Doppler shifts, so you won't see any funny time skips.
  8. Nov 21, 2015 #7


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    You may find yourself thinking that ther
    This is the classic twin paradox, and the explanation is somewhat different than the explanation of the time dilation that you were asking about earlier in the thread. There's an excellent explanation at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
  9. Nov 21, 2015 #8

    Mister T

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    That's precisely why the word paradox appears in the title "Twin Paradox". But it's misnamed because it's not a paradox at all, just an apparent paradox.

    There are several ways to understand the solution to the apparent paradox. Perhaps the simplest is to give a valid argument for ignoring the point of view of the traveler. That's easy. The traveler is not an inertial observer. He has to turn around and this breaks the symmetry between him and the person who stays at home. During the turn-around the traveler is not in a state of inertial motion.

    A deeper and more satisfying way to understand it is to look at it from the point of view of the traveling twin. When he switches inertial reference frames to return home the clocks on Earth that were formerly receding from him are now approaching him. This causes a huge shift in the traveler's reckoning of the reading on the Earth clocks. It's an effect known as the relativity of simultaneity.

    There are other ways to see the resolution of the apparent paradox, such as to introduce a third observer who co-moves with the traveler on his outbound journey. But instead of turning around he continues in the same direction at the same speed so that he is an inertial observer the entire time. During the traveler's return he will see the traveler recede from him at a speed even greater than the speed at which he sees Earth recede from him. Thus he will observe the traveler's clock running even slower than Earth clocks. He will conclude that clocks aboard the ship will record a greater time spent returning than what was spent during the outbound half of the journey. Thus, clocks on the traveler's ship will be behind Earth clocks when the ship arrives back home.
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